A circular, 20-turn coil of radius 5.0 cm is oriented in such a way that its axis makes a 30º angle with a uniform magnetic field of 0.15 T. What is the torque on the coil when it carries a current of 2.5 A?

Question

Quizplus · Accepted Answer

The torque on a coil is given by τ = nIABsinθ, where n is the number of turns, I is the current, A is the area of the coil, B is the magnetic field, and θ is the angle between the axis of the coil and the magnetic field. For this problem, n = 20, I = 2.5 A, A = π(0.05 m)^2, B = 0.15 T, and θ = 30º. Calculating gives τ = 20 * 2.5 * π * (0.05)^2 * 0.15 * sin(30º) = 2.9 × 10^-2 N · m.

A Circular, 20-Turn Coil of Radius 5